\(\int \frac {1+(-1+2 x-e^x x) \log (x)}{x \log (x)} \, dx\) [9600]

Optimal result

Integrand size = 24, antiderivative size = 17 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=-13-e^x+2 x-\log (x)+\log (\log (x)) \]

[Out]

2*x-13-ln(x)-exp(x)+ln(ln(x))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6874, 2225, 45, 2339, 29} \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 x-e^x-\log (x)+\log (\log (x)) \]

[In]

Int[(1 + (-1 + 2*x - E^x*x)*Log[x])/(x*Log[x]),x]

[Out]

-E^x + 2*x - Log[x] + Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^x+\frac {1-\log (x)+2 x \log (x)}{x \log (x)}\right ) \, dx \\ & = -\int e^x \, dx+\int \frac {1-\log (x)+2 x \log (x)}{x \log (x)} \, dx \\ & = -e^x+\int \left (\frac {-1+2 x}{x}+\frac {1}{x \log (x)}\right ) \, dx \\ & = -e^x+\int \frac {-1+2 x}{x} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -e^x+\int \left (2-\frac {1}{x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -e^x+2 x-\log (x)+\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=-e^x+2 x-\log (x)+\log (\log (x)) \]

[In]

Integrate[(1 + (-1 + 2*x - E^x*x)*Log[x])/(x*Log[x]),x]

[Out]

-E^x + 2*x - Log[x] + Log[Log[x]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\)	\(16\)
norman	\(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\)	\(16\)
risch	\(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\)	\(16\)
parallelrisch	\(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\)	\(16\)
parts	\(\ln \left (\ln \left (x \right )\right )+2 x -\ln \left (x \right )-{\mathrm e}^{x}\)	\(16\)

[In]

int(((-exp(x)*x+2*x-1)*ln(x)+1)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))+2*x-ln(x)-exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-exp(x)*x+2*x-1)*log(x)+1)/x/log(x),x, algorithm="fricas")

[Out]

2*x - e^x - log(x) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 x - e^{x} - \log {\left (x \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((-exp(x)*x+2*x-1)*ln(x)+1)/x/ln(x),x)

[Out]

2*x - exp(x) - log(x) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-exp(x)*x+2*x-1)*log(x)+1)/x/log(x),x, algorithm="maxima")

[Out]

2*x - e^x - log(x) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2 \, x - e^{x} - \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-exp(x)*x+2*x-1)*log(x)+1)/x/log(x),x, algorithm="giac")

[Out]

2*x - e^x - log(x) + log(log(x))

Mupad [B] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1+\left (-1+2 x-e^x x\right ) \log (x)}{x \log (x)} \, dx=2\,x+\ln \left (\ln \left (x\right )\right )-{\mathrm {e}}^x-\ln \left (x\right ) \]

[In]

int(-(log(x)*(x*exp(x) - 2*x + 1) - 1)/(x*log(x)),x)

[Out]

2*x + log(log(x)) - exp(x) - log(x)